11-Lect-Pressure Drop in Reactors0

Evaluate 1 w 1 2 dw fa 0 dx 23 b 1 3 krtca 0

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Unformatted text preview: dX 2 2 kCA 0 (1 − X ) [1 − αW ] € € F dX [1 − αW ] dW = 2 A 0 2 kCA 0 (1 − X ) Integrating gives: ȹ α ȹ F X υ X W ȹ 1 − W ȹ = A20 =0 ȹ 2 Ⱥ kCA 0 (1 − X ) kCA 0 (1 − X ) € When ε = 0, we can solve the pressure drop DFQ by separation of variables, substitute y into the dW/dF DFQ and solve. Solving for W or X gives: Ⱥ kC W ȹ α ȹȺ kC W ȹ α ȹ X = A 0 ȹ 1 − W ȹ 1 + Ⱥ A 0 ȹ 1 − W ȹȺ ȹ 2 Ⱥ υ0 € Ⱥ υ 0 ȹ 2 ȺȺ Pressure Drop in Reactors W= 1 − {1 − [(2υ 0α ) kCA 0 ][ X (1 − X )]} 1/ 2 α Lecture 11 - Page 13 Chemical Reaction Engineering - Musgrave € € 5. Evaluate dW = FA 0 dX 2/3 Ⱥ ȹ b ȹ Ⱥ 1/ 3 ȹ Θ B − X ȹ Ⱥ Ⱥ Ⱥ Ⱥ (1 − X ) P ȺC ȹ a Ⱥ P Ⱥ kRTȺCA 0 Ⱥ (1 + εX ) P0 Ⱥ Ⱥ A 0 (1 + εX ) P0 Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ Ⱥ εX Negligible y= α≡ P 1/ 2 = [1 − αW ] P0 2β 0 If ε is zero or εX negligible,then: € (1 − φ ) Ac ρc P0 € dW = kRT FA 0 dX 2/3 P b ȹ 1 / 3 ȹ CA 0 (1 − X ) ȹ Θ B − X ȹ ȹ P0 a Ⱥ € only valid for εX ~0 And using the definition of y: € dW = kRTCA 0 [1 − αW ] FA 0 dX 1/ 2 (1 − X ) 1/ 3 2/3 ȹ b ȹ ȹ Θ B − X ȹ ȹ a Ⱥ Separating variables: € Pressure Drop in Reactors [1 − αW ] 1/ 2 dW = FA 0 dX 2/3 b ȹ 1 / 3 ȹ kRTCA 0 (1 − X ) ȹ Θ B − X ȹ ȹ a Ⱥ Lecture 11 - Page 14 Chemical Reaction Engineering - Musgrave € 7 5. Evaluate [1 − αW ] 1/ 2 dW = FA 0 dX 2/3 b ȹ 1 / 3 ȹ kRTCA 0 (1 − X ) ȹ Θ B − X ȹ ȹ a Ⱥ For A+ 1/2B → C we have b = 1/2, and ΘB is given as 1/2 so: € [1 − αW ] Integrating gives: 1/ 2 dW = 2 FA 0 dX kRTCA 0 (1 − X ) € 2 2 FA 0 32 1 − αW ) = ln(1 − X ) 2( α kRTCA 0 Solving for W or X gives: Ⱥ €1 1 Ⱥ α 2 FA 0 W = − Ⱥ ln(1 − X )Ⱥ α α Ⱥ kRTCA 0 Ⱥ Pressure Drop in Reactors 23 Ⱥ kRTCA 0 3 2 Ⱥ X = 1 − expȺ (1 − αW ) Ⱥ FA 0α 2 Ⱥ Ⱥ Lecture 11 - Page 15 Note: ε is -1/2 and εX is negligible only for small X! Chemical Reaction...
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This note was uploaded on 04/25/2010 for the course CHEN 4330 taught by Professor Staff during the Spring '08 term at Colorado.

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